Question

If $u=f\left(x^3\right)$, $v=g\left(x^2\right)$, $f^{{}^{\prime }}\left(x\right)=\cos x$ and $g^{{}^{\prime }}\left(x\right)=\sin x$ then $\frac{du}{dv}$ is

• A. $\frac{3}{2}x.\cos x^3.cosec{x}^2$
• B. $\frac{2}{3}\sin x^3.\sec x^2$
• C. $\tan x$
• D. none of these

Answer 1

The correct option is A $\frac{3}{2}x.\cos x^3.cosec{x}^2$

$f\left(x^3\right)=u\Rightarrow f^{\prime }\left(x^3\right)\times {3x}^2=\frac{du}{dx}$

$g\left(x^2\right)=v\Rightarrow g^{\prime }\left(x^2\right)\times 2x=\frac{dv}{dx}$

$\therefore \frac{du}{dv}=\frac{3x^2{f}^{\prime }\left(x^3\right)}{2x{g}^{\prime }\left(x^2\right)}=\frac{3x\cos \left(x^3\right)}{2\sin x^2}$

$=\frac{3x}{2}\cos x^3\sec x^2$